olymmath

CMO 2006

Canadian Mathematical Olympiad 2006.

Paper

Problem 1

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Let f(n,k)f(n, k) be the number of ways of distributing kk candies to nn children so that each child receives at most 2 candies. For example, if n=3n=3, then f(3,7)=0,f(3,6)=1f(3,7)=0, f(3,6)=1 and f(3,4)=6f(3,4)=6.

Determine the value of

f(2006,1)+f(2006,4)+f(2006,7)+⋯+f(2006,1000)+f(2006,1003).f(2006,1)+f(2006,4)+f(2006,7)+\cdots+f(2006,1000)+f(2006,1003) .

Comment. Unfortunately, there was an error in the statement of this problem. It was intended that the sum should continue to f(2006,4012)f(2006,4012).

Answer, solutions and source

Problem 2

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Let ABCA B C be an acute-angled triangle. Inscribe a rectangle DEFGD E F G in this triangle so that DD is on AB,EA B, E is on ACA C and both FF and GG are on BCB C. Describe the locus of (i.e., the curve occupied by) the intersections of the diagonals of all possible rectangles DEFGD E F G.

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Problem 3

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In a rectangular array of nonnegative real numbers with mm rows and nn columns, each row and each column contains at least one positive element. Moreover, if a row and a column intersect in a positive element, then the sums of their elements are the same. Prove that m=nm=n.

Answer, solutions and source

Problem 4

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Consider a round-robin tournament with 2n+12 n+1 teams, where each team plays each other team exactly once. We say that three teams X,YX, Y and ZZ, form a cycle triplet if XX beats Y,YY, Y beats ZZ, and ZZ beats XX. There are no ties.

(a) Determine the minimum number of cycle triplets possible.

(b) Determine the maximum number of cycle triplets possible.

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Problem 5

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The vertices of a right triangle ABCA B C inscribed in a circle divide the circumference into three arcs. The right angle is at AA, so that the opposite arc⁡BC\operatorname{arc} B C is a semicircle while arc⁡AB\operatorname{arc} A B and arc⁡AC\operatorname{arc} A C are supplementary. To each of the three arcs, we draw a tangent such that its point of tangency is the midpoint of that portion of the tangent intercepted by the extended lines ABA B and ACA C. More precisely, the point DD on arc⁡BC\operatorname{arc} B C is the midpoint of the segment joining the points D′D^{\prime} and D′′D^{\prime \prime} where the tangent at DD intersects the extended lines ABA B and ACA C. Similarly for EE on arc ACA C and FF on arc ABA B.

Prove that triangle DEFD E F is equilateral.

Answer, solutions and source